A) 0
B) -1
C) 2
D) 1
Correct Answer: D
Solution :
[d]: Let\[I=\int\limits_{1}^{{{e}^{17/2}}}{\frac{\pi \cos (\pi log\,x)}{x}dx}\] Put \[\pi \log x=z\Rightarrow \frac{\pi }{x}dx=dz\] Since, \[1\le x\le {{e}^{17/2}}\Rightarrow 0\le z\le \frac{17\pi }{2}\] \[\therefore \]\[I=\int\limits_{0}^{17\pi /2}{\cos zdz=[sinz]_{0}^{17\pi /2}}\] \[=\sin \frac{17\pi }{2}-\sin 0=\sin \left( 8\pi +\frac{\pi }{2} \right)=\sin \frac{\pi }{2}=1\]
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